# The Plucker relations are sufficient

Consider the Grassmannian of codimension-$d$ subspaces of a given vector space $E$ (over an arbitrary field), which I will define as $$\operatorname{Gr}^d(E) = \{\text{linear surjections } \sigma: E \to F \mid \text{F is any d-dimensional space} \}/\sim$$ where I identify two surjections $(\sigma_1: E \to F_1) \sim (\sigma_2: E \to F_2)$ if there is an isomorphism $m: F_1 \to F_2$ such that $m \sigma_1 = \sigma_2$. Then an isomorphism class of $\sigma$ is determined by the $\ker \sigma$. We also have projective space $\mathbb{P}(E) = \operatorname{Gr}^1(E)$, so that $\operatorname{Sym}^\bullet(E)$ gives homogeneous coordinates on $\mathbb{P}(E)$.

The Plucker embedding is the map $\operatorname{Gr}^d(E) \to \mathbb{P}(\bigwedge^d E)$, taking a map $(\sigma: E \to F)$ to its $d$th exterior power $\bigwedge^d \sigma: \bigwedge^d E \to \bigwedge^d F$. Then $\bigwedge^d \sigma$ is a surjection from $\bigwedge^d E$ to a one-dimensional space, and so lives in $\mathbb{P}(\bigwedge^d E)$. So I am thinking of $\bigwedge^d \sigma$ as a point in the embedding of the Grassmannian, and elements of $\bigwedge^d(E)$ give me the linear coordinate functions. For example, if I write out the matrix of $\sigma: E \to F$ in some basis $(e_1, \ldots, e_m)$ for $E$ and any basis for $F$, then $(\bigwedge^d \sigma)(e_1 \wedge \cdots \wedge e_d)$ is proportional to the determinant of minor where we take the first $d$ columns of the matrix for $\sigma$.

Of course, the question arises that, given some $(s: \bigwedge^d E \to L) \in \mathbb{P}(\bigwedge^d E)$, is $s$ in the image of the Plucker embedding? (i.e., is $s$ a $d$th wedge power?). The answer to this is given by the Plucker relations, which I will state as follows. Define a linear map $\omega^d_E$ on pure tensors by \begin{aligned}\omega^d_E: \bigwedge^{d+1} E \otimes \bigwedge^{d-1} E &\to \bigwedge^d E \otimes \bigwedge^d E\\ v_1 \wedge \cdots \wedge v_{d+1} \otimes u_1 \wedge \cdots \wedge u_{d-1} &\mapsto \sum_{i = 0}^{d+1} v_1 \wedge \cdots \wedge \hat{v_i} \wedge \cdots \wedge v_{d+1} \otimes v_i \wedge u_1 \wedge \cdots \wedge u_{d-1} \end{aligned} Then the point $(s: \bigwedge^d E \to L)$ "satisfies the Plucker relations" if the linear map $(s \otimes s) \circ \omega^d_E$ is zero. From this standpoint, it is lovely to see the necessary condition: if $(\sigma: E \to F) \in \operatorname{Gr}^d(E)$, then $\wedge^d \sigma$ must satisfy the Plucker relations, since the wedge power essentially "pulls through" the $\omega$: $$(\wedge^d \sigma \otimes \wedge^d \sigma) \circ \omega^d_E = \omega_F^d \circ (\wedge^{d+1} \sigma \otimes \wedge^{d-1} \sigma)$$ and of course $\wedge^{d+1} \sigma = 0$ because $F$ is $d$-dimensional. However, I am having trouble with the opposite direction.

How can I see that if $(s: \wedge^d E \to L) \in \mathbb{P}(\bigwedge^d E)$ satisfies the Plucker relations, then it is (proportional to) the $d$th power of some map $\sigma: E \to F$? I would really like a constructive proof: produce some $\sigma: E \to F$ from $s$, and show that as long as $s$ satisfies the Plucker relations, that $\wedge^d \sigma \sim s$.

I can prove similar results for decomposable tensors in $\wedge^d E$, but for some reason with how things are set up, "dualising" that proof is eluding me. There is a proof similar to the style I am thinking of on page 175 of Martin Brandenburg's Thesis, unfortunately I cannot follow the proof starting from the "two short exact sequences". Which leads me to another question:

Are there any good references for the Plucker embedding or Plucker relations thought of in this way?

• What exactly don't you understand in Brandenburg's proof? In the two exact sequences, the left-most map of the first row takes $a\otimes b$ to $a\otimes s(b)-b\otimes s(a)$ and the left-most map of the bottom is just the natural map $\ker t\otimes \bigwedge^{d-1}\to \bigwedge E$, twisted by $L$. (Note that $\ker t = \bigwedge^{d+1}E\otimes L^\vee$ by construction.) That the two diagonal compositions vanish imply that the right-most square can be completed by an isomorphism. Twisting by $L^\vee$ again gives exactly what you want. – Ben May 4 '18 at 17:03
• @Ben Literally everything you just said was not clear to me. How do you know that the upper left map takes $a \otimes b$ to $a \otimes s(b) - b \otimes s(a)$ rather than $a \otimes s(b)$ for example? Also where can I find this result about diagonals vanishing in two short exact sequences? I guess the other thing is that this proof seems entirely pulled-out-of-a-hat, and I'm trying to get some intuition for exactly what is going on. I understand why defining $t$ as the cokernel of a certain map should be right, but I cannot make any sense of the proof from there. – Joppy May 5 '18 at 0:46
• I see. As for intuition, I don’t know, but I will try to explain the details there a bit further later. – Ben May 5 '18 at 6:37
• @Ben Thanks, that would be much appreciated. – Joppy May 5 '18 at 6:41

## 2 Answers

As promised in the comment, here are some more details to Martin Brandenburg's proof. To get an understanding of what is going on, we first put ourselves in the situation that $s = \wedge^d\sigma$ for some surjective $\sigma\colon E\to F$. Why would we want to consider the map $t\colon \bigwedge\nolimits^{d+1}E\to E\otimes\bigwedge\nolimits^{d}F$ defined as \begin{align*}t(v_0\wedge\dots\wedge v_d) &= \sum_{k=0}^d(-1)^kv_k\otimes s(v_0\wedge\dots\wedge\widehat{v_k}\wedge\dots\wedge v_d) \\&=\sum_{k=0}^d(-1)^kv_k\otimes \sigma(v_0)\wedge\dots\wedge\widehat{\sigma(v_k)}\wedge\dots\wedge \sigma(v_d)\;\;? \end{align*} Well, I claim that its image is the kernel of the map $E\otimes\bigwedge^dF\xrightarrow{\sigma\otimes \mathrm{id}_{\wedge^d F}} F\otimes\bigwedge^dF$; since we can reconstruct $\sigma\colon E\to F$ up to isomorphism from its kernel, this map is very much relevant for what we are trying to do.

Let's prove the claim: It is easy to verify that $(\sigma\otimes \mathrm{id}_{\wedge^dF})\circ t$ factorises as $$\bigwedge\nolimits^{d+1}E\xrightarrow{\wedge^{d+1}\sigma}\bigwedge\nolimits^{d+1}F\to F\otimes\bigwedge\nolimits^{d}F,$$ where the latter map sends $v_0\wedge\dots\wedge v_d$ to $\sum_{k=0}^d(-1)^kv_k\otimes v_0\wedge\dots\wedge\widehat{v_k}\wedge\dots\wedge v_d$. But $\bigwedge\nolimits^{d+1}F = 0$; thus, $(\sigma\otimes\mathrm{id}_{\wedge^dF})\circ t = 0$, so that $\mathrm{im}(t)\subset \ker(\sigma\otimes \mathrm{id}_{\wedge^dF})$. Conversely, if $\sigma(w) = 0$, then $t(w\wedge v_1\wedge\dots\wedge v_d)=w\otimes s(v_1\wedge\dots\wedge v_d)$ and so $t$ maps surjectively onto $\ker(\sigma)\otimes\bigwedge^dF = \ker(\sigma\otimes \mathrm{id}_{\wedge^dF})$, as claimed.

What this means is that we have found a reasonable candidate for an inverse of the map $$\left\{E\xrightarrow{\sigma} F\to 0\right\}\to\left\{\bigwedge\nolimits^{d}E\xrightarrow{s}L\to 0\,\middle|\,\text{sat. Plücker}\right\},\sigma\mapsto \wedge^d\sigma,$$ by mapping $s$ to the cokernel of $T_s\colon \bigwedge\nolimits^{d+1}E\otimes L^\vee\xrightarrow{t\otimes \mathrm{id}_{L}}E\otimes L\otimes L^\vee\to E$, where the last map is just the natural isomorphism. The above shows that if we start with a $\sigma$, pass to $\wedge^d\sigma$, and then take the cokernel of $T_{\wedge^d\sigma}$, we get back $\sigma$ up to isomorphism. It remains to show that starting with some $s$ satisfying the Plücker relations, the candidate-inverse is well-defined (i.e., that the cokernel has rank $d$,) and that if we pass to the cokernel $\sigma\colon E\to F:=\mathrm{coker}(T_s)$ and then apply $\wedge^d$, we get back $s$ up to isomorphism. The latter is what those exact sequences are for, but we can phrase it without them:

We have two quotients of $\bigwedge^dE\otimes L$, namely, $\wedge^d\sigma\otimes \mathrm{id}_{L}\colon \bigwedge^dE\otimes L\to \bigwedge^dF\otimes L$ and $s\otimes \mathrm{id}_{L}\colon\bigwedge^dE\otimes L\to L\otimes L$ and we aim to show that they are isomorphic as quotients, i.e., that $\ker(\wedge^d\sigma\otimes\mathrm{id}_{L}) = \ker(s\otimes \mathrm{id}_{L})$. For this, we give nice presentations of those kernels.

For one, since $\ker(\sigma\otimes\mathrm{id}_{L})$ is the image of $t$, the kernel of $\wedge^d\sigma\otimes\mathrm{id}_{L}$ is the image of the map $\alpha\colon \bigwedge^{d-1}E\otimes\bigwedge^{d+1}E\to \bigwedge^{d}E\otimes L$, mapping $v\otimes w$ to $v\wedge t(w)$.

For the other map, note that $\ker(s\otimes \mathrm{id}_{L})=\ker(s)\otimes L$ is generated by elements of the form $v\otimes s(w)-w\otimes s(v)$, since, for $s(v)=0$ and $f = s(w)\in L$ arbitrary, $v\otimes s(w) - w\otimes s(v) = v\otimes f$. In particular, with $\beta\colon \bigwedge^dE\otimes\bigwedge^dE\to \bigwedge^dE\otimes L$ mapping $v\otimes w$ to $v\otimes s(w)- w\otimes s(v)$, we get $\ker(s\otimes \mathrm{id}_{L}) = \mathrm{im}{(\beta)}$.

Thus, if we manage to show that $(s\otimes \mathrm{id}_{L})\circ\alpha = 0$ and $(\wedge^d\sigma\otimes\mathrm{id}_{L})\circ\beta = 0$, then we conclude $$\ker(\wedge^d\sigma\otimes\mathrm{id}_{L})=\mathrm{im}{(\alpha)}\subset\ker(s\otimes \mathrm{id}_{L}) = \mathrm{im}{(\beta)}\subset \ker(\wedge^d\sigma\otimes\mathrm{id}_{L}),$$ which implies equality everywhere. In particular, $L\cong\bigwedge^dF$ as quotients of $\bigwedge^d E$ and so $\bigwedge^dF$ is invertible, hence $F$ has rank $d$; this is all we wanted to show.

Finally, we show the two identities $(s\otimes \mathrm{id}_{L})\circ\alpha = 0$ and $(\wedge^d\sigma\otimes\mathrm{id}_{L})\circ\beta = 0$. Tracing through the definitions shows that the former is the Plücker relation, and that the second is equivalent to $\wedge^d\sigma v\otimes s(w) = \wedge^d\sigma w\otimes s(v)$ for all $v,w\in\bigwedge^dE$. That is, we want $\wedge^d\sigma\otimes s$ to be symmetric. By construction of $\sigma$, we always have $$0 = \sum_{k=0}^d(-1)^{k}\sigma w_k\otimes s(w_0\wedge\dots\wedge \widehat{w_{k}}\wedge \dots\wedge w_d),$$ and so the symmetry of $\wedge^d\sigma\otimes s$ follows from what M. Brandenburg calls the Symmetry Lemma (4.4.15); I have nothing to add to his proof of this lemma.

• Thanks for taking the time to explain! I still don't understand how you go from the fact that the kernel of $\sigma \otimes \mathrm{id}_L$ is the image of $t$, and arrive at the conclusion that the kernel of $\wedge^d \sigma \otimes \mathrm{id}_L$ is precisely the image of $\alpha$. – Joppy May 8 '18 at 1:27
• This holds quite generally: if $\varphi\colon V\to W$ is surjective, then so is $\wedge^d \varphi$ and the kernel is generated by the elements of the form $v_1\wedge\dots\wedge v_d$ for all $v_1\in\ker\varphi$ and $v_2,\dots,v_d\in V$. I will try to find a reference or a quick argument later in the day. – Ben May 8 '18 at 5:31
• Ah, I see - I definitely believe that. I think I'm beginning to see how to connect this to the usual decomposability of vectors in $\wedge^d E$ results. – Joppy May 8 '18 at 6:29
• Here is a simple argument: If $\wedge^d\varphi(v_1,\dots,v_d) = 0$, then the $\varphi(v_i)$ are linearly dependent. Thus, there exist scalars $\lambda_i$ such that $\varphi(\sum_i\lambda_i v_i) = \sum_i\lambda_i\varphi(v_i) = 0$, hence, $v_0 := \sum_i\lambda_i v_i\in\ker(\varphi)$. Say $\lambda_1 = 1$, possibly after renumbering and rescaling. Then $v_1\wedge\dots\wedge v_d = v_0\wedge v_2\wedge\dots\wedge v_d - (v_0-v_1)\wedge v_2\wedge\dots\wedge v_d$ where the first term has a factor in $\ker(\varphi)$ and the second vanishes. It remains to show that it suffices to consider those elements.. – Ben May 8 '18 at 7:05

I cannot put this as a comment and I'm sorry that it is not a complete answer, but a good introduction to Grassmanninans is written by Gathmann: http://www.mathematik.uni-kl.de/~gathmann/class/alggeom-2014/alggeom-2014-c8.pdf

In particular the answer to your second question could be Corollary 8.13 (look at the proof).

• Thanks for the reference, but this isn't what I'm looking for. In those notes, the relations given are the vanishing of some $(n-1+k) \times (n-1+k)$: this is a relation of degree $n-1+k$. The Plucker relations I gave above are different, and in particular are always degree 2. – Joppy Apr 29 '18 at 1:36